Let \(A\) be the subset of \(\mathbb{R}^2\) consisting of the union of the topologist’s sine curve and the broken-line path from \((0, -1)\) to \((0, -2)\) to \((1, -2)\) to \((1, \sin 1)\). See Figure 61.4. We call \(A\) the closed topologist’s sine curve. Show that if \(C\) is a subspace of \(S^2\) homeomorphic to the closed topologist’s sine curve, then \(C\) separates \(S^2\).
Solution
1. The closed topologist’s sine curve \(A\) is a simple closed curve (a Jordan curve) in \(\mathbb{R}^2\).
\(A\) is the union of the topologist’s sine curve \(\{(x, \sin(1/x)) : 0 < x \le 1\}\) with the broken-line path from \((0,-1)\) to \((0,-2)\) to \((1,-2)\) to \((1, \sin 1)\), together with the limiting segment \(\{0\} \times [-1, 1]\); this forms a simple closed curve (a homeomorphic image of \(S^1\)).
2. By the Jordan curve theorem, a simple closed curve in \(S^2\) separates \(S^2\) into two components.
the Jordan curve theorem.
3. Since \(C\) is homeomorphic to \(A\) (a simple closed curve), \(C\) is itself a simple closed curve in \(S^2\).
hypothesis and 1.
4. Hence \(C\) separates \(S^2\).
2 and 3.
5. Q.E.D.
4.